I have a question regarding the simulation of a GBM. I have found similar questions here but nothing which takes reference to my specific problem:

Given a GBM of the form

$dS(t) = \mu S(t) dt + \sigma S(t) dW(t)$

it is clear that this SDE has a closed form solution in

$S(t) = S(0) exp ([\mu - \frac{1}{2}\sigma^2]t + \sigma W(t))$

for a given $S(0)$.

Now, I have found sources claiming that in order to simulate the whole trajectory of the GBM, one needs to convert it to its discrete form (e.g., a similar question here or Iacus: "Simulation and Inference for Stochastic Differential Equations", 62f.). Yet, in Glasserman: "Monte Carlo Methods in Fin. Eng.", p. 94, I find that

That makes sense, thanks a lot (and sorry about the typos). So, do I understand correctly then that the discrete form is usually an approximation and used if there is no closed form (i.e., no exact analytical solution)? And by approaching the limit of 0 with my timestep, I can overcome the problem (somehwat) of an approximation error?
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gu7zJan 28 '13 at 13:24

@gu7z you can accept the answer if you're satisfied by it!
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SRKX♦Jan 28 '13 at 13:38

One can view his equation as the Euler scheme for log(S). In that case discretization and exact soluation have no difference! The reason why one has to go in discrete steps in the way to generate W(t) from i.i.d. random variables.
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Christian FriesJan 28 '13 at 20:09

is the exact solution of the SDE. Hence, the discretization is exact (which is a special case here).

Note that $W(t_{i+1})$ is not independent of $W(t_{i})$ but $W(t_{i+1})-W(t_{i})$ is independet from $W(t_{i})-W(t_{i-1})$. So in order to simulate the discrete points $S(t_{j})$ for different $j$ you use the representation above with i.i.d. random variable $Z_{j}$ with $W(t_{j})-W(t_{j-1}) = \sqrt{t_{j}-t_{j-1}} Z_{j}$ and not the representation

Thank's for pointing out the typos and sorry about that. I follow your point, just one question: Why does the time dependent part in the stochastic element drop out? should sigma Z_j not be dependent on the time step?
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gu7zJan 28 '13 at 13:30

@gu7z not it's not, by definition of the model.
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SRKX♦Jan 28 '13 at 13:39

1

@gu7z Yes. In the formulation $W(t_j)−W(t_{j−1}) = Z_j$ the $Z_j$ is a normal distributed random variable with mean zero and variance $t_{j}-t_{j-1}$. But since I wrote that the Z_j are i.i.d. we have indeed to rescale them and write $W(t_j)−W(t_{j−1}) = \sqrt{t_{j}-t_{j-1}} Z_j$ (or otherwise assume equidistributed time stepping. I corrected that in my post!
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Christian FriesJan 28 '13 at 20:05